A pendulum is made of two rods that are firmly welded together to make a cross. The rod AB is 2 m long with a mass of 20 kg. The rod CD is 1 m long with a mass of 10 kg. Recall that for a rod the moment of inertia about its center of mass is given by IG = 1/12 m L2 cross2 What is the moment of inertia about A for member AB?

 IA = 80/3 kgm^2

Explanation:

Given:-

- The mass of rod AB, m1 = 20 kg

- The length of rod AB, L1 = 2m

- The mass of rod CD, m2 = 10 kg

- The length of rod CD, L2 = 1m

Find:-

What is the moment of inertia about A for member AB?

Solution:-

- The moment of inertia About point "O" the center of rod AB is given as:

                               IG = 1/12*m1*L^2

- To shift the axis of moment of inertia for any object at a distance "d" from the center of mass of that particular object we apply the parallel axis theorem. The new moment of inertia about any arbitrary point, which in our case A end of rod AB is:

                              IA = IG + m1*d^2

- Where the distance "d" from center of rod AB to its ends is 1/2*L1 = 1 m.

So the moment of inertia for rod AB at point A would be:

                              IA = 1/12*m1*L^2 + m1*0.5*L1^2

                              IA = 1/3 * m1*L1^2

                              IA = 1/3*20*2^2

                             IA = 80/3 kgm^2